If $P_A(B)>P_{\neg A}(B)$ holds, then does $P_B(A) > P_{\neg B}(A)$?

Question

I tried to prove the statement;

If $P_A(B)>P_{\neg A}(B)$ holds, then does $P_B(A) > P_{\neg B}(A)$?

※Here $P_A(B)$ stands for the conditional probability in which $B$ occurs when $A$ occurs.

But I couldn't.

So my question is "how to prove it(if the statement is true...)"?

or is there a counterexample?

Answer 1

$$\frac{P(A\cap B)}{P(A)}=P(B|A)>P(B|\bar{A})=\frac{P(\bar A\cap B)}{P(\bar A)}$$ is equivalent to

$$P(A\cap B)P(\bar A)>P(\bar{A}\cap B)P(A).$$ Now, having in mind,

$$P(\bar A)=1-P(A),\quad P(\bar A\cap B)=P(B)-P(A\cap B)$$ we get

$$P(A\cap B)(1-P(A)>(P(B)-P(A\cap B))P(A).$$ In other words

$$P(A\cap B)-P(A)P(A\cap B)>P(A)P(B)-P(A)P(A\cap B).$$ That is

$$P(A\cap B)>P(A)P(B).$$ So, we have shown

$$P(B|A)>P(B|\bar{A})\iff P(A\cap B)>P(A)P(B).$$

Because of the symmetry of the role of $A,B$ we are done.

Answer 2

This is true. Write down all conditional expectations and use $P(B\cap A^c)=P(B)-P(A\cap B)$,$P(A\cap B^c)=P(A)-P(A \cap B)$. You will see that the hypothesis is same as the conclusion.